Let
R0 be the initial resistance of both conductors
∴ At temperature
θ their resistance will be,
R1=R0(1+α1θ) and
R2=R0(1+α2θ)
for, series combination,
Rs=R1+R2Rs0(1+αsθ)=R_0\left(1+\alpha_1 \theta\right)
+R0(1+α2θ)
where
Rs0=R0+R0=2R0 \therefore \quad 2
R0(1+αsθ)=2R0+R0 \theta\left(\alpha_1+\alpha_2\right)
or\quad \alpha_s=\frac{\alpha_1+\alpha_2}{2}
forparallelcombination,Rp=\frac{R_1 R_2}{R_1+R_2}
where,R_{p 0}=\frac{R_0 R_0}{R_0+R_0}=\frac{R_0}{2} \therefore \quad \frac{\mathrm{R}_0}{2}
(1+αpθ)=R0(2+α1θ+α2θ)R02(1+α1θ+α2θ+α1α2θ)
as
α1 and
α2 are small quantities
∴α1α2 is negligible
or
αp=2+(α1+α2)θα1+α2=2α1+α2[1−(α1+α2)θ]
as
(α1+α2)2 is negligible
∴αp=2α1+α2